As written in the title I want to show that $f$ is continuous at the point $x_0=1$ using the $\epsilon$-$\delta$ definition.
Here's my attempt, but I am not sure it's correct.
Let $\epsilon>0$ and define $\delta=\epsilon$, then $\forall x$ sucht that |$x-x_0$|$ \leq \delta$ we have
|$f(x)-f(x_0)$|$=$|$\frac{\sqrt{2x}}{\sqrt{x+1}}-1$|$=$|$\sqrt{\frac{2x}{x+1}}-1$|$=$|$\frac{(\sqrt{\frac{2x}{x+1}}-1)(\sqrt{\frac{2x}{x+1}}+1)}{(\sqrt{\frac{2x}{x+1}}+1)}$|$=$|$\frac{\frac{x-1}{x+1}}{(\sqrt{\frac{2x}{x+1}}+1)}$|$=$$\frac{|x-1|}{|(x+1)(\sqrt{\frac{2x}{x+1}}+1)|}$
Now since the denominator is greater than one, I get
$\frac{|x-1|}{|(x+1)(\sqrt{\frac{2x}{x+1}}+1)|} \leq |x-1| \leq \delta = \epsilon$
Hence, $f$ is continuous in the point $x_0=1$.
Thank you for your help